Mixing
Why is Convolution Well-Defined (Simple Example)
$begingroup$
I am having trouble understanding why convolution is well-defined.
Let's take a simple example:
$(Omega, mathcal{F}, P)$ probability space and $X_{1}, X_{2}$ two real random variables where $P(X_{1}=3)=frac{1}{2},P(X_{2}=2)=frac{1}{4}$
And $P(X_{1}=1)=frac{1}{5},P(X_{2}=4)=frac{1}{3}$
Then my understanding of convolution is
$P_{X_{1}+X_{2}}circ A^{-1}$ where $A: X_{1} times X_{2}to mathbb R,A(x_{1},x_{2})=x_{1}+x_{2}$
So surely, if, for instance $X_{1}+X_{2}=5$, I get more than one preimage, and hence how can convolution be well-defined?
In the above case, I would get:
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(3)P_{X_{2}}(2)=frac{1}{2}timesfrac{1}{4}=frac{1}{8}$ while
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(1)P_{X_{2}}(4)=frac{1}{5}timesfrac{1}{3}=frac{1}{15}$
I do not know where I am going wrong in my understanding of convolution. Any help is greatly appreciated.
probability probability-theory random-variables convolution
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding why convolution is well-defined.
Let's take a simple example:
$(Omega, mathcal{F}, P)$ probability space and $X_{1}, X_{2}$ two real random variables where $P(X_{1}=3)=frac{1}{2},P(X_{2}=2)=frac{1}{4}$
And $P(X_{1}=1)=frac{1}{5},P(X_{2}=4)=frac{1}{3}$
Then my understanding of convolution is
$P_{X_{1}+X_{2}}circ A^{-1}$ where $A: X_{1} times X_{2}to mathbb R,A(x_{1},x_{2})=x_{1}+x_{2}$
So surely, if, for instance $X_{1}+X_{2}=5$, I get more than one preimage, and hence how can convolution be well-defined?
In the above case, I would get:
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(3)P_{X_{2}}(2)=frac{1}{2}timesfrac{1}{4}=frac{1}{8}$ while
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(1)P_{X_{2}}(4)=frac{1}{5}timesfrac{1}{3}=frac{1}{15}$
I do not know where I am going wrong in my understanding of convolution. Any help is greatly appreciated.
probability probability-theory random-variables convolution
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding why convolution is well-defined.
Let's take a simple example:
$(Omega, mathcal{F}, P)$ probability space and $X_{1}, X_{2}$ two real random variables where $P(X_{1}=3)=frac{1}{2},P(X_{2}=2)=frac{1}{4}$
And $P(X_{1}=1)=frac{1}{5},P(X_{2}=4)=frac{1}{3}$
Then my understanding of convolution is
$P_{X_{1}+X_{2}}circ A^{-1}$ where $A: X_{1} times X_{2}to mathbb R,A(x_{1},x_{2})=x_{1}+x_{2}$
So surely, if, for instance $X_{1}+X_{2}=5$, I get more than one preimage, and hence how can convolution be well-defined?
In the above case, I would get:
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(3)P_{X_{2}}(2)=frac{1}{2}timesfrac{1}{4}=frac{1}{8}$ while
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(1)P_{X_{2}}(4)=frac{1}{5}timesfrac{1}{3}=frac{1}{15}$
I do not know where I am going wrong in my understanding of convolution. Any help is greatly appreciated.
probability probability-theory random-variables convolution
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
$endgroup$
I am having trouble understanding why convolution is well-defined.
Let's take a simple example:
$(Omega, mathcal{F}, P)$ probability space and $X_{1}, X_{2}$ two real random variables where $P(X_{1}=3)=frac{1}{2},P(X_{2}=2)=frac{1}{4}$
And $P(X_{1}=1)=frac{1}{5},P(X_{2}=4)=frac{1}{3}$
Then my understanding of convolution is
$P_{X_{1}+X_{2}}circ A^{-1}$ where $A: X_{1} times X_{2}to mathbb R,A(x_{1},x_{2})=x_{1}+x_{2}$
So surely, if, for instance $X_{1}+X_{2}=5$, I get more than one preimage, and hence how can convolution be well-defined?
In the above case, I would get:
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(3)P_{X_{2}}(2)=frac{1}{2}timesfrac{1}{4}=frac{1}{8}$ while
$P_{X_{1}+X_{2}}circ A^{-1}(5)=P_{X_{1}}(1)P_{X_{2}}(4)=frac{1}{5}timesfrac{1}{3}=frac{1}{15}$
I do not know where I am going wrong in my understanding of convolution. Any help is greatly appreciated.
probability probability-theory random-variables convolution
probability probability-theory random-variables convolution
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
asked Jan 28 at 14:32
MinaThumaMinaThuma
1968
1968
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $P$ denotes the probability measure on $(Omega,mathcal F)$ then it induces for every random variable $Z$ a probability measure $P_Z$ on $(mathbb R,mathcal B)$ that is prescribed by:$$Bmapsto P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Here ${Zin B}$ abbreviates ${omegainOmegamid Z(omega)in B}$ and $P(Zin B)$ abbreviates $P({Zin B})$
So we have $P_Z(B)=P({Zin B}$ for Borel subsets of $mathbb R$.
Another notation of this probability is $PZ^{-1}$ prescribed by:$$Bmapsto P(Z^{-1}(B))=P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Observe that $P_Z$ and $PZ^{-1}$ are notations for the same measure.
Also random vector $(X_1,X_2)$ induces a probability measure.
This time denoted as $P_{(X_1,X_2)}$ and defined on $(mathbb R^2,mathcal B^2)$.
If $A:mathbb R^2tomathbb R$ is prescribed by $(x,y)mapsto x+y$ then $A$ is a Borel-measurable function.
That means that it can be looked at as a random variable on space $(mathbb R^2,mathcal B^2,P_{(X_1,X_2)})$.
Applying the principle that was mentioned above on space $(mathbb R^2,mathcal B,P_{(X_1,X_2)})$ we have measure $P_{(X_1,X_2)}A^{-1}$ on $(mathbb R,mathcal B)$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{Acirc(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$
In your question you mix up the two notations, and this can be a source of confusion on its own.
If $X_1,X_2$ are random variables then so is $X_1+X_2$.
This with e.g.:$$P_{X_1+X_2}({5})=P(X_1+X_2in{5})=P(X_1+X_2=5)$$
If moreover $X_1,X_2$ only take integers as value then this can be expanded to:$$cdots=sum_{n,minmathbb Zwedge n+m=5}P(X_1=nwedge X_2=m)$$
answered Jan 28 at 15:09
drhabdrhab
104k545136
$endgroup$
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090927%2fwhy-is-convolution-well-defined-simple-example%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $P$ denotes the probability measure on $(Omega,mathcal F)$ then it induces for every random variable $Z$ a probability measure $P_Z$ on $(mathbb R,mathcal B)$ that is prescribed by:$$Bmapsto P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Here ${Zin B}$ abbreviates ${omegainOmegamid Z(omega)in B}$ and $P(Zin B)$ abbreviates $P({Zin B})$
So we have $P_Z(B)=P({Zin B}$ for Borel subsets of $mathbb R$.
Another notation of this probability is $PZ^{-1}$ prescribed by:$$Bmapsto P(Z^{-1}(B))=P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Observe that $P_Z$ and $PZ^{-1}$ are notations for the same measure.
Also random vector $(X_1,X_2)$ induces a probability measure.
This time denoted as $P_{(X_1,X_2)}$ and defined on $(mathbb R^2,mathcal B^2)$.
If $A:mathbb R^2tomathbb R$ is prescribed by $(x,y)mapsto x+y$ then $A$ is a Borel-measurable function.
That means that it can be looked at as a random variable on space $(mathbb R^2,mathcal B^2,P_{(X_1,X_2)})$.
Applying the principle that was mentioned above on space $(mathbb R^2,mathcal B,P_{(X_1,X_2)})$ we have measure $P_{(X_1,X_2)}A^{-1}$ on $(mathbb R,mathcal B)$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{Acirc(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$
In your question you mix up the two notations, and this can be a source of confusion on its own.
If $X_1,X_2$ are random variables then so is $X_1+X_2$.
This with e.g.:$$P_{X_1+X_2}({5})=P(X_1+X_2in{5})=P(X_1+X_2=5)$$
If moreover $X_1,X_2$ only take integers as value then this can be expanded to:$$cdots=sum_{n,minmathbb Zwedge n+m=5}P(X_1=nwedge X_2=m)$$
answered Jan 28 at 15:09
drhabdrhab
104k545136
$endgroup$
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
add a comment |
$begingroup$
If $P$ denotes the probability measure on $(Omega,mathcal F)$ then it induces for every random variable $Z$ a probability measure $P_Z$ on $(mathbb R,mathcal B)$ that is prescribed by:$$Bmapsto P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Here ${Zin B}$ abbreviates ${omegainOmegamid Z(omega)in B}$ and $P(Zin B)$ abbreviates $P({Zin B})$
So we have $P_Z(B)=P({Zin B}$ for Borel subsets of $mathbb R$.
Another notation of this probability is $PZ^{-1}$ prescribed by:$$Bmapsto P(Z^{-1}(B))=P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Observe that $P_Z$ and $PZ^{-1}$ are notations for the same measure.
Also random vector $(X_1,X_2)$ induces a probability measure.
This time denoted as $P_{(X_1,X_2)}$ and defined on $(mathbb R^2,mathcal B^2)$.
If $A:mathbb R^2tomathbb R$ is prescribed by $(x,y)mapsto x+y$ then $A$ is a Borel-measurable function.
That means that it can be looked at as a random variable on space $(mathbb R^2,mathcal B^2,P_{(X_1,X_2)})$.
Applying the principle that was mentioned above on space $(mathbb R^2,mathcal B,P_{(X_1,X_2)})$ we have measure $P_{(X_1,X_2)}A^{-1}$ on $(mathbb R,mathcal B)$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{Acirc(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$
In your question you mix up the two notations, and this can be a source of confusion on its own.
If $X_1,X_2$ are random variables then so is $X_1+X_2$.
This with e.g.:$$P_{X_1+X_2}({5})=P(X_1+X_2in{5})=P(X_1+X_2=5)$$
If moreover $X_1,X_2$ only take integers as value then this can be expanded to:$$cdots=sum_{n,minmathbb Zwedge n+m=5}P(X_1=nwedge X_2=m)$$
answered Jan 28 at 15:09
drhabdrhab
104k545136
$endgroup$
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
add a comment |
$begingroup$
If $P$ denotes the probability measure on $(Omega,mathcal F)$ then it induces for every random variable $Z$ a probability measure $P_Z$ on $(mathbb R,mathcal B)$ that is prescribed by:$$Bmapsto P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Here ${Zin B}$ abbreviates ${omegainOmegamid Z(omega)in B}$ and $P(Zin B)$ abbreviates $P({Zin B})$
So we have $P_Z(B)=P({Zin B}$ for Borel subsets of $mathbb R$.
Another notation of this probability is $PZ^{-1}$ prescribed by:$$Bmapsto P(Z^{-1}(B))=P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Observe that $P_Z$ and $PZ^{-1}$ are notations for the same measure.
Also random vector $(X_1,X_2)$ induces a probability measure.
This time denoted as $P_{(X_1,X_2)}$ and defined on $(mathbb R^2,mathcal B^2)$.
If $A:mathbb R^2tomathbb R$ is prescribed by $(x,y)mapsto x+y$ then $A$ is a Borel-measurable function.
That means that it can be looked at as a random variable on space $(mathbb R^2,mathcal B^2,P_{(X_1,X_2)})$.
Applying the principle that was mentioned above on space $(mathbb R^2,mathcal B,P_{(X_1,X_2)})$ we have measure $P_{(X_1,X_2)}A^{-1}$ on $(mathbb R,mathcal B)$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{Acirc(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$
In your question you mix up the two notations, and this can be a source of confusion on its own.
If $X_1,X_2$ are random variables then so is $X_1+X_2$.
This with e.g.:$$P_{X_1+X_2}({5})=P(X_1+X_2in{5})=P(X_1+X_2=5)$$
If moreover $X_1,X_2$ only take integers as value then this can be expanded to:$$cdots=sum_{n,minmathbb Zwedge n+m=5}P(X_1=nwedge X_2=m)$$
answered Jan 28 at 15:09
drhabdrhab
104k545136
$endgroup$
If $P$ denotes the probability measure on $(Omega,mathcal F)$ then it induces for every random variable $Z$ a probability measure $P_Z$ on $(mathbb R,mathcal B)$ that is prescribed by:$$Bmapsto P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Here ${Zin B}$ abbreviates ${omegainOmegamid Z(omega)in B}$ and $P(Zin B)$ abbreviates $P({Zin B})$
So we have $P_Z(B)=P({Zin B}$ for Borel subsets of $mathbb R$.
Another notation of this probability is $PZ^{-1}$ prescribed by:$$Bmapsto P(Z^{-1}(B))=P({omegainOmegamid Z(omega)in B})=P({Zin B})=P(Zin B)$$
Observe that $P_Z$ and $PZ^{-1}$ are notations for the same measure.
Also random vector $(X_1,X_2)$ induces a probability measure.
This time denoted as $P_{(X_1,X_2)}$ and defined on $(mathbb R^2,mathcal B^2)$.
If $A:mathbb R^2tomathbb R$ is prescribed by $(x,y)mapsto x+y$ then $A$ is a Borel-measurable function.
That means that it can be looked at as a random variable on space $(mathbb R^2,mathcal B^2,P_{(X_1,X_2)})$.
Applying the principle that was mentioned above on space $(mathbb R^2,mathcal B,P_{(X_1,X_2)})$ we have measure $P_{(X_1,X_2)}A^{-1}$ on $(mathbb R,mathcal B)$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{Acirc(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$
In your question you mix up the two notations, and this can be a source of confusion on its own.
If $X_1,X_2$ are random variables then so is $X_1+X_2$.
This with e.g.:$$P_{X_1+X_2}({5})=P(X_1+X_2in{5})=P(X_1+X_2=5)$$
If moreover $X_1,X_2$ only take integers as value then this can be expanded to:$$cdots=sum_{n,minmathbb Zwedge n+m=5}P(X_1=nwedge X_2=m)$$
answered Jan 28 at 15:09
drhabdrhab
104k545136
answered Jan 28 at 15:09
drhabdrhab
104k545136
answered Jan 28 at 15:09
drhabdrhab
104k545136
answered Jan 28 at 15:09
drhabdrhab
104k545136
104k545136
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
add a comment |
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $sum_{n,m in mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$
$endgroup$
– MinaThuma
Jan 28 at 17:51
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
$begingroup$
Yes, that is correct.
$endgroup$
– drhab
Jan 28 at 18:57
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090927%2fwhy-is-convolution-well-defined-simple-example%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
